Question

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving with a speed v_i. After the collision, the orange disk moves along a direction that makes an angle θ with its initial direction of motion. The velocities of the two disks are perpendicular after the collision. Determine the final speed of each disk. (Use any variable or symbol stated above as necessary.)

Answer 1

Using Momentum conservation, in vertical and horizontal direction:

In Horizontal direction:

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

u1x = Speed of orange disk before collision = vi m/sec

u2x = Speed of yellow disk before collision = 0 m/sec

v1x = Speed of orange disk after collision = v1*cos

Since both disk are perpendicular after collision, So yellow disk will be at (90 - ) with the initial direction of motion below positive x-axis.

v2x = Speed of yellow disk after collision = v1*cos (90 - ) = v1*sin

m1 = m2 = m

So,

u1x + u2x = v1x + v2x

vi = v1*cos + v2*sin ............. eq 1

Now In vertical direction

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

u1y = 0 m/sec & u2y = 0 m/sec

v1y = v1*sin

v2y = -v2*sin (90 - ) = -v2*cos

m1 = m2

So,

0 + 0 = v1*sin - v2*cos

v2 = v1*(sin )/(cos ) = v1*tan

v2 = v1*tan

Using above value in eq 1:

vi = v1*cos + v2*sin

vi = v1*cos + v1*tan *sin

vi = v1*[cos + sin *tan ]

vi = v1*[cos + sin *sin /cos ]

vi = v1*[cos² + sin²]/cos

Since [cos² + sin²] = 1

vi = v1/cos

v1 = speed of orange disk after collision = vi*cos

v2 = v1*tan

v2 = vi*cos *tan

v2 = vi*cos *sin /cos

v2 = speed of yellow disk after collision = vi*sin

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